[leetcode] 310. Minimum Height Trees

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return [3, 4]

 

Solution:

使用二維數組構造圖G[N][N]，數組G[i]表示節點i的全部相鄰節點。數組dist[i]記錄節點i的度。

一、根據輸入構造圖G和數組dist

二、把度位1的節點放入BFS隊列Q

三、刪除全部度爲1的節點，並更新和這些節點相鄰的節點的度。當度爲1時繼續加入隊列Q。

四、重複3， 直到剩餘節點數<=2

 1 vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) 
 2     {
 3         if (n == 1) 
 4             return {0};
 5         vector<int> result;
 6         vector<int> dist(n, 0);
 7         vector<vector<int> > G(n, vector<int>());
 8         queue<int> Q;
 9         
10         // init graph
11         for (auto elem : edges)
12         {
13             G[elem.first].push_back(elem.second);
14             G[elem.second].push_back(elem.first);
15             dist[elem.second]++;
16             dist[elem.first]++;
17         }
18         for (int node = 0; node < dist.size(); node++)
19         {
20             if (dist[node] == 1)
21                 Q.push(node);
22         }
23         
24         int count = n;
25         while (count > 2)
26         {
27             int qsize = Q.size();
28             for (int i = 0; i < qsize; i++)
29             {
30                 int node = Q.front();
31                 Q.pop();
32                 count--;
33                 for (auto elem : G[node])
34                 {
35                     dist[elem]--;
36                     if (dist[elem] == 1)
37                         Q.push(elem);
38                 }
39             }
40         }
41         
42         while (!Q.empty())
43         {
44             result.push_back(Q.front());
45             Q.pop();
46         }
47         
48         return result;
49     }